Ncert Exercises & Solutions

7.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

${NaNO}_{2}$ is the salt of a strong base (NaOH) and a weak acid ( ${HNO}_{2}$ ).

${NO}_{2}^{-} + H_{2} O \leftrightarrow {HNO}_{2} + {OH}^{-}$
$K_{h} = \frac{[{HNO}_{2}] [{OH}^{-}]}{[{NO}_{2}^{-}]}$
$\Rightarrow \frac{K_{w}}{K_{α}} = \frac{10^{- 14}}{4.5 \times 10^{- 4}} = . 22 \times 10^{- 10}$

Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species present in the solution will be:

$[{NO}_{2}^{-}] = . 04 - x; 0.04$
$[{HNO}_{2}] = x$
$[{OH}^{-}] = x$
$K_{h} = \frac{x^{2}}{0.04} = 0.22 \times 10^{- 10}$
$x^{2} = . 0088 \times 10^{10}$
$x = 0.93 \times 10^{- 5}$
$∴ [{OH}^{-}] = 0.093 \times 10^{- 5} M$
$[H_{2} O^{+}] = \frac{10^{- 14}}{. 093 \times 10^{- 5}} = 10.75 \times 10^{- 9} M$
$\Rightarrow pH = - \log (10.75 \times 10^{- 9})$
$= 7.96$
$Therefore, degree of hydrolysis$
$= \frac{x}{0.04} = \frac{. 093 \times 10^{- 5}}{. 04}$
$= 2.325 \times 10^{- 5}$

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